# Let #y = f(x)# be a twice-differentiable function such that #f(1) = 2# and . What is the value of #(d^2y)/(dx^2)# at # x = 1#?

#(d^2y)/(dx^2) = 28# when#x=1#

We have:

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To find the value of (\frac{{d^2y}}{{dx^2}}) at (x = 1), we need to differentiate the function (f(x)) twice with respect to (x) and then evaluate it at (x = 1).

Given (y = f(x)), and (f(1) = 2), we need to find (f''(1)), the second derivative of (f(x)) evaluated at (x = 1).

We're provided with (f'(x) = 3x^2 - 6x + 2).

Differentiating (f'(x)) with respect to (x) again, we get: [\frac{{d^2y}}{{dx^2}} = \frac{{d}}{{dx}}(3x^2 - 6x + 2) = 6x - 6.]

Substitute (x = 1) into (f''(x)): [f''(1) = 6(1) - 6 = 6 - 6 = 0.]

So, the value of (\frac{{d^2y}}{{dx^2}}) at (x = 1) is (0).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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